Simplify the following expression and state the condition under which the simplification is valid. $x = \dfrac{9n^2 - 54n - 63}{-9n^2 - 27n + 630}$
Explanation: First factor out the greatest common factors in the numerator and in the denominator. $ x = \dfrac {9(n^2 - 6n - 7)} {-9(n^2 + 3n - 70)} $ $ x = -\dfrac{9}{9} \cdot \dfrac{n^2 - 6n - 7}{n^2 + 3n - 70} $ Simplify: $ x = - \dfrac{n^2 - 6n - 7}{n^2 + 3n - 70}$ Next factor the numerator and denominator. $ x = - \dfrac{(n - 7)(n + 1)}{(n - 7)(n + 10)}$ Assuming $n \neq 7$ , we can cancel the $n - 7$ $ x = - \dfrac{n + 1}{n + 10}$ Therefore: $ x = \dfrac{ -n - 1 }{ n + 10 }$, $n \neq 7$